Solution:
Induction on $n$. We claim that we can find $N$ with $n$ digits, all $1$ or $2$, so that $N$ is divisible by $2^{n}$.

True for $n = 1$: take $N = 2$.

Suppose it is true for $n$. If $2^{n + 1}$ divides $N$, then since $2^{n + 1}$ divides $2 \times 10^{n}$, it also divides $N'$ obtained from $N$ by placing a $2$ in front of it.

If $2^{n + 1}$ does not divide $N$, then $N = 2^{n} \times \text{odd}$ and $10^{n} = 2^{n} \times \text{odd}$, so $N + 10^{n}$ (in other words, the $n + 1$ digit number obtained by placing a $1$ in front of $N$) is divisible by $2^{n + 1}$.